Exercise 143 Page 576

y= ax^2+ bx+ cLet's start by identifying the values of a, b, and c. y=2x^2-12x+6 ⇕ y= 2x^2+( - 12)x+ 6 We can see that a= 2, b= - 12, and c= 6. Recall that, if a>0 the parabola opens upwards. Conversely, if a<0 the parabola opens downwards.

In the given function we have a= 2, which is greater than 0. Thus, the parabola opens upwards and we will have a minimum value. The vertex is always the lowest or the highest point on the graph. Therefore, in this case the vertex represents the minimum value of the function.

Vertex: ( - b/2a, f( - b/2a ) )Since we already know the values of a and b, we can substitute them into the formula for the x-coordinate of the vertex.

We have found that the x-coordinate of the vertex is x=3. To find the y-coordinate, we need to substitute 3 for x in the given equation.

We found the y-coordinate, and now we know that the vertex is (3,- 12).

1. Find the axis of symmetry.
2. Identify the y-intercept and its reflection across the axis of symmetry.
3. Connect the points and vertex with a parabola.Let's go through these steps one at a time.

   Finding the Axis of Symmetry

   The axis of symmetry is a vertical line with equation x=- b2 a. Note that the formula for the axis of symmetry is the same as the formula for the x-coordinate of the vertex. Recall that in Part B we found that the x-coordinate of the vertex equals x=3. Thus, the axis of symmetry is also 3.

   Identifying the y-intercept and its Reflection

   The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,6). Let's plot this point and its reflection across the axis of symmetry.

   

   Connecting the Points

   We can now draw the graph of the function. Since a=2, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.